\(P=\frac{x-y}{x+y}\)\(\Rightarrow P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}\)

Ta có: \(2x^2+2y^2=5xy\left(1\right)\)

\(\Leftrightarrow2x^2-4xy+2y^2=xy\)

\(\Leftrightarrow2\left(x-y\right)^2=xy\)

\(\Leftrightarrow\left(x-y\right)^2=\frac{xy}{2}\)(2)

Từ \(\left(1\right)\Rightarrow2x^2+4xy+2y^2=9xy\)

\(\Leftrightarrow2\left(x+y\right)^2=9xy\)

\(\Leftrightarrow\left(x+y\right)^2=\frac{9xy}{2}\)(3)

Thay (2)và (3) vào \(P^2\)ta được:

\(P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{xy}{2}.\frac{2}{9xy}=\frac{1}{9}\)

Nhận thấy \(y>x>0\Rightarrow x-y< 0\Rightarrow\frac{x-y}{x+y}< 0\Rightarrow P< 0\)

\(\Rightarrow P=\frac{-1}{3}\)

E= \(\frac{3x^2-8x+6}{x^2-2x+1}\)= \(\frac{3x^2-8x+6}{\left(x-1\right)^2}\)

Đặt t= x-1 => x= t+1

E= \(\frac{3\left(t+1\right)^2-8\left(t+1\right)+6}{t^2}\)= \(\frac{3\left(t^2+2t+1\right)-8t-8+6}{t^2}\)=\(\frac{3t^2+6t+3-8t-8+6}{t^2}\)=\(\frac{3t^2-2t+1}{t^2}\)= \(\frac{3t^2}{t^2}\)\(-\frac{2t}{t^2}\)\(\frac{1}{t^2}\)= 3-\(\frac{2}{t}\)+\(\frac{1}{t^2}\)

Đặt \(\frac{1}{t}\)= b \(\Rightarrow\)E= b²-2b+3= b²-2b+1+2= (b-1)²+2\(\ge\)2

Min E= 2 khi b-1= 0 \(\Rightarrow\)b= 1\(\Rightarrow\)\(\frac{1}{t}\)= 1\(\Rightarrow\)\(\frac{1}{x-1}\)= 1 \(\Leftrightarrow\)x-1= 1 \(\Rightarrow\)x=1 

 \(x+y=4\Rightarrow\frac{x+y}{2}=2\Rightarrow\sqrt{\frac{x+y}{2}}=\sqrt{2}\)

\(P.\sqrt{\frac{x+y}{2}}=\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}\)

\(\Leftrightarrow\sqrt{2}P=\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}\)

\(\Leftrightarrow\sqrt{2}P\ge x+\frac{1}{x}+y+\frac{1}{y}\)

\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge4-3x\)

\(y+\frac{1}{y}=\left(\frac{1}{y}+4y\right)-3y\ge4-3y\)

\(\Rightarrow\sqrt{2}P\ge8-3\left(x+y\right)=8-3.4=-4\)

đến đay sau răng

Vì \(EI//AB\)và \(AB//DM\Rightarrow EI//DM\)  nên \(\frac{AE}{AD}=\frac{AI}{IM}=\frac{EI}{DM}\left(1\right)\)

Vì \(AB//MC\) và \(KF//AB\Rightarrow KF//MC\Rightarrow\frac{BK}{BM}=\frac{BF}{BC}=\frac{KF}{MC}\left(2\right)\)

Ta có : IK // AB \(\Rightarrow\frac{AI}{AM}=\frac{BK}{BM}\left(3\right)\)

Từ(1) (2)(3), \(\Rightarrow\frac{EI}{DM}=\frac{KF}{MC}\)

Mà DM=MC => EI=KF \(\left(\cdot\right)\)

Ta có : \(IK//DM\Rightarrow\frac{IK}{DM}=\frac{IB}{BD}=\frac{BK}{BM}\left(4\right)\)

Ta có : EI // AB

\(\Rightarrow\frac{IB}{BD}=\frac{AE}{AD}\left(5\right)\)

Từ (1)(4)(5) \(\Rightarrow\frac{EI}{DM}=\frac{IK}{DM}\Rightarrow EI=IK\left(\cdot\cdot\right)\)

Từ \(\left(\cdot\right)\left(\cdot\cdot\right)\Rightarrow EI=IK=KF\)

1. Cho B=\(\frac{x^3}{x^2-4}\)− \(\frac{x}{x-2}\)− \(\frac{2}{x+2}\)Rút gọn B và tìm \(x\inℕ\)để b là số nguyên

2. Cho \(x\ge0;\) \(x\ne0;\) \(x\ne9;\). Tìm X để \(\left(1-\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\cdot\)\(\left(\frac{1}{1-\sqrt{x}}+\frac{2}{\sqrt{x}-3}\right)=2\)

Gợi ý: \(\left(y=\sqrt{x}\right)\Rightarrow x=y^2\)

3.Cho B=\(\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x-3}}\)rút gọn B và tìm X để \(\frac{1}{B}\le-\frac{5}{2}\)

GIÚP MÌNH NHA!!! 5 Tick 1 câu

a)\(\frac{1-3x}{2}\)-\(\frac{x+3}{2}\) b)\(\frac{2\left(x+y\right)\left(x-y\right)}{x}\)-\(\frac{-2y^2}{x}\)

c)\(\frac{3x+1}{x+y}\)-\(\frac{2x-3}{x+y}\)

d)\(\frac{xy}{2x-y}\)-\(\frac{x^2-1}{y-2x}\)

e)\(\frac{4x-1}{3x^2y}\)-\(\frac{7x-1}{3x^2y}\)

giúp em với